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MATH132 Applied mathematics 1A (UKZN) - 2021 Tutorials R75,00   Add to cart

Exam (elaborations)

MATH132 Applied mathematics 1A (UKZN) - 2021 Tutorials

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This document includes all 12 sets of tutorial questions from 2021 for the course MATH132 Applied Mathematics 1A at UKZN and will allow you to gain plenty of experience in the types of questions you can expect in tests and exams for this module.

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  • February 7, 2022
  • 25
  • 2021/2022
  • Exam (elaborations)
  • Answers
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APPLIED MATHEMATICS 1A (MATH132)
ADDITIONAL PROBLEMS 1

1. Suppose that the augmented matrix for a system of linear equations has been completed
reduced to the following matrix
 
1 −6 0 0 3 −2
 0 0 1 0 4 7 
 .
 0 0 0 1 5 8 
0 0 0 0 0 0
Write down the general solution of the linear system. (Ans: x1 = −2 + 6x2 − 3x5 , x2
is free, x3 = 7 − 4x5 , x4 = 8 − 5x5 , x5 is free)
2. Do the three planes 2x2 − 2x3 = 1, x1 + 2x2 − x3 = −3 and x1 − x2 + 2x3 = 3 have at
least one common point of intersection? Explain.
3. Consider the system
x1 + x2 + 2x3 = a
x1 + x3 = b
2x1 + x2 + 3x3 = c
where a, b and c are arbitrary real numbers. For the system to be consistent what
condition must a, b and c satisfy? (Ans: c = a + b)
4. Show that if ad − bc ̸= 0, then the system
ax1 + bx2 = r
cx1 + dx2 = s
has a unique solution.
5. Find all values of k for which the linear system
x1 + 2x2 + x3 = 3
2x1 + 3x2 + 3x3 = 4
3x1 + 3x2 + (k 2 − 3)x3 = k
has (a) no solution, (b) a unique solution and (c) infinitely many solutions.
(Ans: (a) k = −3, (b) k ̸= ±3, (c) k = 3)
6. (a) A system of linear equations with fewer equations than unknowns is often called an
underdetermined system. Suppose that such a system is consistent. Explain why
it must have an infinite number of solutions.
(b) A system of linear equations with more equations than unknowns is often called
an overdetermined system. Show that the following overdetermined system is
consistent.
x1 − 2x2 = 3
3x1 − x2 = 14
x1 − 7x2 = −2

,7. A system of linear equations is called homogeneous if the right hand side of each equa-
tion is zero. Such a system always has at least one solution, i.e. x1 = x2 = · · · = xn = 0.
This is called the trivial solution. All other solutions are called nontrivial solutions.
For what values of k do the following homogeneous systems have nontrivial solutions?
x 1 − x2 + x 3 = 0
(k − 2)x1 + x2 = 0
(a) (b) 3x1 + x2 + 4x3 = 0
x1 + (k − 2)x2 = 0
4x1 + 12x2 + kx3 = 0
(Ans: (a) k = 1 or k = 3, (b) k = 8)

8. (a) Solve the system

x1 + 2x2 + x3 = 3
−3x1 − x2 + 2x3 = 1
5x2 + 3x3 = 2

using the Gauss–Jordan elimination method (i.e. transforming the augmented ma-
trix to reduced row echelon form). (Ans: (x1 , x2 , x3 ) = (3, −2, 4))
 
−2
(b) Use your answer in part (a) to write the vector  1  as a linear combination of
      −5
1 1 3
the vectors −3 , 2 and 1 .
    
0 3 2

, APPLIED MATHEMATICS 1A (MATH132)
ADDITIONAL PROBLEMS 2

       
1 2 4 3
1. Let v1 =  0 , v2 = 1 , v3 = 2 and w = 1 .
     
−1 3 6 2
(a) How many vectors are in {v1 , v2 , v3 }? Is w in {v1 , v2 , v3 }?
(b) How many vectors are in Span {v1 , v2 , v3 }?
(c) Is w in Span {v1 , v2 , v3 }? Why? (Ans: Yes, since w can be expressed as a linear
combination of v1 , v2 and v3 )
   
1 3 −1 2
2. Consider matrix A = 2 5 −1 and vector b = 1 .
  
2 8 −2 3
(a) Show that b is a linear combination of the column vectors a1 , a2 and a3 of A.
(b) List three other vectors in Span {a1 , a2 , a3 }.
     
2 4 x
   
3. Let v1 = −1 and v2 = 1 be vectors in R . Furthermore let v =
3  y  be a
4 6 z
vector in R such that v ∈ Span {v1 , v2 }. Find the equation that describes Span {v1 , v2 }.
3

(Ans: 5x − 2y − 3z = 0)
     
4 1 2
4. Determine the value of c so that  c  is in the plane spanned by  1  and  −1 .
−3 −2 5
(Ans: 7/3)
   
1 −2 3 8
5. Let A =  2 −3 2 and b = 7 . Solve the equation Ax = b. Use your answer
 
−1 1 2 4
to solve the nonlinear system

xy − 2 y + 3yz = 8

2xy − 3 y + 2yz = 7

−xy + y + 2yz = 4

for x, y and z. (Ans: x = 5/9, y = 9, z = 1/3)
     
2 1 1
6. Let u =  1 , v =  3  and w =  −7 . Show that 2u − 3v − w = 0. Hence
1 2 −4
(without using row operations) find x1 and x2 that satisfy the equation
   
1 2 [ ] 1
 −7 1  x1 =  3  .
x2
−4 1 2

(Ans: x1 = −1/3, x2 = 2/3)

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